The integral turns out to be doubly bad. The behaviour when $x$ is large is the more obvious badness. We show that there is also fatal badness at $1$, by showing that $\displaystyle\int_1^2 \dfrac{x^6\,dx}{x^6-1}$ diverges.

Note that $x^6-1=(x-1)(x^5+x^4+x^3+x^2+x+1)$. When $x\ge 1$, each of the terms $x^5,x^4,x^3,x^2,x, 1$ is $\le x^6$. It follows that if $\epsilon\gt 0$, then
$$I_\epsilon=\int_{1+\epsilon}^2 \dfrac{x^6\,dx}{x^6-1}\gt \int_{1+\epsilon}^2 \frac{1}{6}\cdot \frac{dx}{x-1}.$$

The change of variable $u=x-1$ shows that
$$I_\epsilon\gt \frac{1}{6}\int_\epsilon^1\frac{du}{u}.$$
But it is a familiar fact that $\displaystyle\int_\epsilon^1\dfrac{du}{u}$ blows up as $\epsilon\to 0^+$.